Solve the inequality
\[\left| \frac{2x - 1}{x - 1} \right| > 2.\]
Explanation: From the given inequality, either $\frac{2x - 1}{x - 1} > 2$ or $\frac{2x - 1}{x - 1} < -2.$

The inequality $\frac{2x - 1}{x - 1} > 2$ becomes
\[\frac{2x - 1}{x - 1} - 2 > 0,\]or
\[\frac{1}{x - 1} > 0.\]This is satisfied when $x > 1.$

The inequality $\frac{2x - 1}{x - 1} < -2$ becomes
\[\frac{2x - 1}{x - 1} + 2 < 0,\]or
\[\frac{4x - 3}{x - 1} < 0.\]If $x < \frac{3}{4},$ then $4x - 3 < 0$ and $x - 1 < 0,$ so the inequality is not satisfied.

If $\frac{3}{4} < x < 1,$ then $4x - 3 > 0$ and $x - 1 < 0,$ so the inequality is satisfied.

If $x > 1,$ then $4x - 3 > 0$ and $x - 1 > 0,$ so the inequality is not satisfied.

Thus, the solution is
\[x \in \boxed{\left( \frac{3}{4}, 1 \right) \cup (1, \infty)}.\]